10.（2024·邵阳校级月考）传统文化幻方是一种中国传统游戏，它是将从一到若干个数的自然数排成纵横各为若干个数的正方形，使在同一行、同一列和同一对角线上的几个数的和都相等. 类比幻方，我们给出如图所示的方格，要使方格中横向、纵向及对角线方向上的实数相乘的结果都相等，则$A$、$B$、$C$、$D$之和为______.

$3\sqrt{5}+3$ 解析：对角线方向上的实数相乘的结果为$5\sqrt{2}\times\sqrt{10}\times\sqrt{2}=10\sqrt{10}$，根据方格中横向、纵向及对角线方向上的实数相乘的结果都相等得$A\times5\times\sqrt{2}=10\sqrt{10}$，解得$A = 2\sqrt{5}$，$B\times\sqrt{10}\times10=10\sqrt{10}$，解得$B = 1$，$5\times\sqrt{10}\times C=10\sqrt{10}$，解得$C = 2$，$\sqrt{2}\times10\times D=10\sqrt{10}$，解得$D=\sqrt{5}$，$\therefore A$、$B$、$C$、$D$之和为$2\sqrt{5}+1+2+\sqrt{5}=3\sqrt{5}+3$.

11. 计算：
（1）（2023·上海中考）$\sqrt[3]{8}+\frac{1}{2+\sqrt{5}}-\left(\frac{1}{3}\right)^{-2}+|\sqrt{5}-3|$；
（2）$\sqrt{2}\times\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)-\frac{\sqrt{18}-\sqrt{8}}{\sqrt{2}}$；
（3）$\left(\sqrt{27}\times3\sqrt{6}+\frac{4}{5}\sqrt{50}-8\sqrt{\frac{1}{2}}\right)\div\sqrt{2}$；
（4）$\left(\frac{\sqrt{18}}{3}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\frac{3}{\sqrt{3}}+\sqrt{5}\right)$.

(1)原式$=2+\sqrt{5}-2 - 9+3-\sqrt{5}=-6$.
(2)原式$=\sqrt{2}\times(\sqrt{2}+\frac{\sqrt{2}}{2})-\frac{3\sqrt{2}-2\sqrt{2}}{\sqrt{2}}=\sqrt{2}\times\frac{3\sqrt{2}}{2}-\frac{\sqrt{2}}{\sqrt{2}}=3 - 1=2$.
(3)原式$=(3\sqrt{3}\times3\sqrt{6}+\frac{4}{5}\times5\sqrt{2}-8\times\frac{\sqrt{2}}{2})\div\sqrt{2}=(27\sqrt{2}+4\sqrt{2}-4\sqrt{2})\div\sqrt{2}=27\sqrt{2}\div\sqrt{2}=27$.
(4)原式$=[\sqrt{2}+(\sqrt{3}-\sqrt{5})]\times[\sqrt{2}-(\sqrt{3}-\sqrt{5})]=(\sqrt{2})^{2}-(\sqrt{3}-\sqrt{5})^{2}=2-(3 - 2\sqrt{15}+5)=2 - 8+2\sqrt{15}=-6+2\sqrt{15}$.

12. 有个填写运算符号的游戏：在“$\frac{\sqrt{2}}{2}\square\sqrt{8}\square\sqrt{18}\square4\sqrt{2}$”中的每个$\square$内，填入$+$，$-$，$\times$，$\div$中的某一个（可重复使用），然后计算结果.
（1）计算：$\frac{\sqrt{2}}{2}+\sqrt{8}-\sqrt{18}-4\sqrt{2}$；
（2）若$\frac{\sqrt{2}}{2}\div\sqrt{8}\times\sqrt{18}\square4\sqrt{2}=-\frac{13}{4}\sqrt{2}$，请推算$\square$内的符号；
（3）在“$\frac{\sqrt{2}}{2}\square\sqrt{8}\square\sqrt{18}-4\sqrt{2}$”的$\square$内填入符号后，使计算所得结果最大，直接写出这个最大结果.

(1)$\frac{\sqrt{2}}{2}+\sqrt{8}-\sqrt{18}-4\sqrt{2}=\frac{\sqrt{2}}{2}+2\sqrt{2}-3\sqrt{2}-4\sqrt{2}=-\frac{9\sqrt{2}}{2}$.
(2)因为$\frac{\sqrt{2}}{2}\div\sqrt{8}\times\sqrt{18}\square4\sqrt{2}=-\frac{13}{4}\sqrt{2}$，所以$\frac{\sqrt{2}}{2}\times\frac{1}{2\sqrt{2}}\times3\sqrt{2}\square4\sqrt{2}=-\frac{13}{4}\sqrt{2}$，所以$\frac{3\sqrt{2}}{4}\square4\sqrt{2}=-\frac{13}{4}\sqrt{2}$. 因为$\frac{3\sqrt{2}}{4}-4\sqrt{2}=-\frac{13}{4}\sqrt{2}$，所以$\square$内的符号是“-”.
(3)$12-\frac{7\sqrt{2}}{2}$ 解析：$\frac{\sqrt{2}}{2}+\sqrt{8}\times\sqrt{18}-4\sqrt{2}=12-\frac{7\sqrt{2}}{2}$，第一个$\square$内填“+”，第二个$\square$内填“×”可使计算所得结果最大，最大是$12-\frac{7\sqrt{2}}{2}$.

13.（2024·长沙期末）若$\sum_{i = 1}^{k}f(i)=f(1)+f(2)+f(3)+\cdots +f(k)$，则$\left(\sqrt{3n + 2}+\frac{2}{\sqrt{2}}\right)\cdot\sum_{i = 1}^{n}\frac{1}{\sqrt{3i + 2}+\sqrt{3i - 1}}=$（ ）

A. $n$
B. $2n$
C. $\sqrt{3n + 1}$
D. $n\sqrt{2}$

A 解析：根据题意得$\sum_{i = 1}^{k}f(i)=f(1)+f(2)+f(3)+\cdots + f(k)$，
$\therefore\sum_{i = 1}^{n}\frac{1}{\sqrt{3i+2}+\sqrt{3i - 1}}=\frac{1}{\sqrt{3\times1+2}+\sqrt{3\times1 - 1}}+\frac{1}{\sqrt{3\times2+2}+\sqrt{3\times2 - 1}}+\frac{1}{\sqrt{3\times3+2}+\sqrt{3\times3 - 1}}+\cdots+\frac{1}{\sqrt{3n+2}+\sqrt{3n - 1}}=\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{8}+\sqrt{5}}+\frac{1}{\sqrt{11}+\sqrt{8}}+\cdots+\frac{1}{\sqrt{3n+2}+\sqrt{3n - 1}}=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}+\frac{\sqrt{8}-\sqrt{5}}{(\sqrt{8}+\sqrt{5})(\sqrt{8}-\sqrt{5})}+\frac{\sqrt{11}-\sqrt{8}}{(\sqrt{11}+\sqrt{8})(\sqrt{11}-\sqrt{8})}+\cdots+\frac{\sqrt{3n+2}-\sqrt{3n - 1}}{(\sqrt{3n+2}+\sqrt{3n - 1})(\sqrt{3n+2}-\sqrt{3n - 1})}=\frac{\sqrt{5}-\sqrt{2}}{3}+\frac{\sqrt{8}-\sqrt{5}}{3}+\frac{\sqrt{11}-\sqrt{8}}{3}+\cdots+\frac{\sqrt{3n+2}-\sqrt{3n - 1}}{3}=\frac{\sqrt{5}}{3}-\frac{\sqrt{2}}{3}+\frac{\sqrt{8}}{3}-\frac{\sqrt{5}}{3}+\frac{\sqrt{11}}{3}-\frac{\sqrt{8}}{3}+\cdots+\frac{\sqrt{3n+2}}{3}-\frac{\sqrt{3n - 1}}{3}=-\frac{\sqrt{2}}{3}+\frac{\sqrt{3n+2}}{3}$，$\therefore(\sqrt{3n+2}+\frac{2}{\sqrt{2}})\cdot\sum_{i = 1}^{n}\frac{1}{\sqrt{3i+2}+\sqrt{3i - 1}}=(\sqrt{3n+2}+\sqrt{2})(-\frac{\sqrt{2}}{3}+\frac{\sqrt{3n+2}}{3})=\frac{3n+2}{3}+\frac{\sqrt{2}\cdot\sqrt{3n+2}}{3}-\frac{\sqrt{2}\cdot\sqrt{3n+2}}{3}-\frac{2}{3}=\frac{3n}{3}=n$. 故选A.

14.（1）先化简，再求值：$\frac{x\sqrt{y}-y\sqrt{x}}{x\sqrt{y}+y\sqrt{x}}-\frac{x\sqrt{y}+y\sqrt{x}}{y\sqrt{x}-x\sqrt{y}}$，其中$x = 3$，$y = 2$；
（2）先化简，再求值：$\left(\frac{1}{a-\sqrt{ab}}+\frac{1}{\sqrt{ab}+b}\right)\div\frac{\sqrt{ab}}{a - b}$，其中$a=\sqrt{3}+1$，$b=\sqrt{3}-1$.

(1)原式$=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}\sqrt{y}(\sqrt{x}+\sqrt{y})}\cdot\frac{\sqrt{x}\sqrt{y}(\sqrt{x}+\sqrt{y})}{\sqrt{x}\sqrt{y}(\sqrt{y}-\sqrt{x})}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y}-\sqrt{x}}=-\frac{(\sqrt{x}-\sqrt{y})^{2}}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}-\frac{(\sqrt{x}+\sqrt{y})^{2}}{(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})}=\frac{x + y-2\sqrt{xy}}{x - y}-\frac{x + y+2\sqrt{xy}}{y - x}=\frac{x + y-2\sqrt{xy}+x + y+2\sqrt{xy}}{x - y}=\frac{2(x + y)}{x - y}$，
当$x = 3$，$y = 2$时，原式$=\frac{2\times(3 + 2)}{3 - 2}=10$.
(2)原式$=(\frac{1}{a-\sqrt{ab}}+\frac{1}{\sqrt{ab}+b})\cdot\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}=[\frac{1}{\sqrt{a}(\sqrt{a}-\sqrt{b})}+\frac{1}{\sqrt{b}(\sqrt{a}+\sqrt{b})}]\cdot\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}=\frac{1}{\sqrt{a}(\sqrt{a}-\sqrt{b})}\cdot\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}+\frac{1}{\sqrt{b}(\sqrt{a}+\sqrt{b})}\cdot\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}}+\frac{\sqrt{a}-\sqrt{b}}{b\sqrt{a}}=\frac{\sqrt{ab}+b}{ab}+\frac{a-\sqrt{ab}}{ab}=\frac{\sqrt{ab}+b+a-\sqrt{ab}}{ab}=\frac{a + b}{ab}$.$\because a=\sqrt{3}+1$，$b=\sqrt{3}-1$，$\therefore a + b=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}$，$ab=(\sqrt{3}+1)(\sqrt{3}-1)=2$，则$\frac{a + b}{ab}=\frac{2\sqrt{3}}{2}=\sqrt{3}$.